Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z^2 - 12z + 27}{z - 3} \times \dfrac{2z + 14}{-5z^2 + 45z} $
Explanation: First factor the quadratic. $q = \dfrac{(z - 9)(z - 3)}{z - 3} \times \dfrac{2z + 14}{-5z^2 + 45z} $ Then factor out any other terms. $q = \dfrac{(z - 9)(z - 3)}{z - 3} \times \dfrac{2(z + 7)}{-5z(z - 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (z - 9)(z - 3) \times 2(z + 7) } { (z - 3) \times -5z(z - 9) } $ $q = \dfrac{ 2(z - 9)(z - 3)(z + 7)}{ -5z(z - 3)(z - 9)} $ Notice that $(z - 3)$ and $(z - 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 2\cancel{(z - 9)}(z - 3)(z + 7)}{ -5z(z - 3)\cancel{(z - 9)}} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $q = \dfrac{ 2\cancel{(z - 9)}\cancel{(z - 3)}(z + 7)}{ -5z\cancel{(z - 3)}\cancel{(z - 9)}} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $q = \dfrac{2(z + 7)}{-5z} $ $q = \dfrac{-2(z + 7)}{5z} ; \space z \neq 9 ; \space z \neq 3 $